Re: help on sql

"RA" <angani_at_gmail.com> a écrit dans le message de news: db579098-d4b0-4778-8f65-3cff8e43d9a2_at_f20g2000prn.googlegroups.com...
| Hi sql gurus,
|
| I need some help and also curious on how to write this in single sql
| statement, requirements goes like this
|
| Table structure
| ================
| create table find_fixed_open (
| find number,
| fixed number,
| open number,
| bug_when date
| );
|
|
| Data
| =============
| insert into find_fixed_open values(5,3,2,'10/01/2009');
| insert into find_fixed_open values(52,38,16,'10/02/2009');
| insert into find_fixed_open values(68,45,39,'10/03/2009');
| insert into find_fixed_open values(112,59,92,'10/04/2009');
| insert into find_fixed_open values(45,12,125,'10/05/2009');
|
|
| Formula
| =========
| Open = Find - fixed + previous rows open(basically open is cumulative)
| i.e.
| 5 - 3 + 0 = 2
| 52 - 38 + 2 = 16
| 68 - 45 + 16 = 39
| 112 - 59 + 39 = 92
| 45 - 12 + 92 = 125
|
| Please help me and enlighten me.
|
| Thank you
| Raju

SQL> select bug_when, find, fixed, open,   2 sum(find-fixed) over (order by bug_when) computed_open   3 from find_fixed_open
  4 order by bug_when
  5 /
BUG_WHEN FIND FIXED OPEN COMPUTED_OPEN ---------- ---------- ---------- ---------- -------------

10/01/2009          5          3          2             2
10/02/2009         52         38         16            16
10/03/2009         68         45         39            39
10/04/2009        112         59         92            92
10/05/2009         45         12        125           125

5 rows selected.

Regards
Michel Received on Sat Oct 10 2009 - 01:25:25 CDT